[tex]x = 10 \ cm\\p = 2\\szukane:\\y = ?\\f = ?\\\\Rozwiazanie\\\\p=\frac{y}{x} \ \ \rightarrow \ \ y = p\cdot x\\\\y = 2\cdot10 \ cm\\\\y = 20 \ cm[/tex]
Z równania zwierciadła (soczewki):
[tex]\frac{1}{f} = \frac{1}{x} + \frac{1}{y}\\\\\frac{1}{f} = \frac{x+y}{xy}\\\\f = \frac{xy}{x+y} = \frac{10cm\cdot20cm}{10cm+20cm} =\frac{200cm^{2}}{30cm}\\\\f\approx6,7 \ cm \ - \ dlugosc \ ogniskowej[/tex]
